A Direct Sum of Members of a Certain Class of Modules

Question

Let $S$ be a class of $R$-modules and let an $R$-module $M$ be countably generated. Suppose that, for every direct summand $K$ of $M$, each element of $K$ belongs to a direct summand of $K$ that is isomorphic to a member of $S$. I could not prove that "$M$ is isomorphic to a direct sum of members of $S$", though this statement seems true anyhow. I appreciate any help!

Answer 1

Start with listing generators $x_1, x_2, \ldots$. Now by induction we will find submodules $M_1, M_2, \ldots, M_n$ such that each $M_i$ is a member of $S$, the sum $M(n) = M_1 + \cdots + M_n$ is direct in $M$, the module $M(n)$ is a direct summand of $M$, and we have $x_1, \ldots, x_n \in M(n)$.

To start apply the given to $x_1 \in K = M$ as suggested by zcn. This gives $M_1 = M(1)$.

For the induction step, choose a complement $K$ to $M(n)$, i.e., such that $M = M(n) \oplus K$. Let $y \in K$ be the projection of $x_{n + 1}$. Then by assumption we get a direct summand $M_{n + 1}$ of $K$ containing $y$. This works.

Then it is clear that $M = \bigoplus_{i\ge 1} M_i$ because each $x_n$ is in the right hand side.